How to determine if 3 points on a 3-D graph are collinear? Let the points $A, B$ and $C$ be $(x_1, y_1, z_1), (x_2, y_2, z_2)$ and $(x_3, y_3, z_3)$ respectively. How do I prove that the 3 points are collinear? What is the formula? 
 A: From $A(x_1,y_1,z_1),B(x_2,y_2,z_2),C(x_3,y_3,z_3)$ we can get their position vectors.
$\vec{AB}=(x_2-x_1,y_2-y_1,z_2-z_1)$ and $\vec{AC}=(x_3-x_1,y_3-y_1,z_3-z_1)$.
Then $||\vec{AB}\times\vec{AC}||=0\implies A,B,C$ collinear.
A: The three points $A,B,C$ are collinear if and only if there exists a real number $k$ such that
$$x_3-x_1=k(x_2-x_1)\ \ \text{and}\ \ y_3-y_1=k(y_2-y_1)\ \ \text{and}\ \ z_3-z_1=k(z_2-z_1).$$
A: The formula is 
\begin{equation*}
\text{rank} \left(
\begin{array}{cccc}
1 & x_1 & x_2 & x_3\\
1 & y_1 & y_2 & y_3 \\
1 & z_1 & z_2 &z_3
\end{array}
\right)\le 2
\end{equation*}
that is, the following three minors are zero
\begin{equation*}
 \left|
\begin{array}{ccc}
1 & x_1 & x_2 \\
1 & y_1 & y_2 \\
1 & z_1 & z_2 
\end{array}
\right|
=
\left|
\begin{array}{ccc}
1 & x_1 & x_3 \\
1 & y_1 & y_3 \\
1 & z_1 & z_3 
\end{array}
\right|=
\left|
\begin{array}{ccc}
1 & x_2 & x_3 \\
1 & y_2 & y_3 \\
1 & z_2 & z_3 
\end{array}
\right|=0
\end{equation*}
The rank $\le 2$ condition  also works for $3$ points in $n$ dimensions.
A: Find the three distances between the points.
Use Heron's formula (and these three distances) to find the area of the triangle.
If the area is positive, then the three points are not collinear. They form a triangle.
If the area is 0, then the three points are collinear.
A: If the distance between |AB|+|BC|=|AC| then A,B,C are collinear. 
